I know that Γ'(1) = -γ, but how does one prove this?
Starting from the basics, we have that:
Γ(x) = \int_0^\infty e^{-t} t^{x-1} dt
How do we differentiate this? How do we then find that
Γ'(1) = \int_0^\infty e^{-t} \log(t) dt
and how would one solve this integral?
Answer
Taken from this answer:
By the recursive relation \Gamma(x+1)=x\Gamma(x), we get
\small{\log(\Gamma(x))=\log(\Gamma(n+x))-\log(x)-\log(x+1)-\log(x+2)-\dots-\log(x+n-1)}\tag{1}
Differentiating (1) with respect to x, evaluating at x=1, and letting n\to\infty yields
\begin{align} \frac{\Gamma'(1)}{\Gamma(1)}&=\log(n)+O\left(\frac1n\right)-\frac11-\frac12-\frac13-\dots-\frac1n\\ &\to-\gamma\tag{2} \end{align}
A discussion of why \frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\log(x)+O\left(\frac1x\right), and a proof of the log-convexity of \Gamma(x), is given in the answer cited above. It is easy to accept since \log(\Gamma(n+1))=\log(\Gamma(n))+\log(n).
Another Approach
\begin{align} \int_0^\infty\log(t)\,e^{-t}\,\mathrm{d}t &=\lim_{n\to\infty}\int_0^n\log(t)\,\left(1-\frac{t}{n}\right)^n\,\mathrm{d}t\tag{3a}\\ &=\lim_{n\to\infty}n\int_0^1(\log(t)+\log(n))\,(1-t)^n\,\mathrm{d}t\tag{3b}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)+n\int_0^1\log(1-t)\,t^n\,\mathrm{d}t\right)\tag{3c}\\ &=\lim_{n\to\infty}\left(\frac{n}{n+1}\log(n)-\frac{n}{n+1}H_{n+1}\right)\tag{3d}\\[6pt] &=-\gamma\tag{3e} \end{align}
Explanation:
\text{(3a)}: Monotone Convergence and e^{-t}=\lim\limits_{n\to\infty}\left(1-\frac tn\right)^n
\text{(3b)}: substitute t\mapsto nt
\text{(3c)}: substitute t\mapsto1-t
\text{(3d)}: integrate by parts \int_0^1\log(1-t)\,t^n\,\mathrm{d}t=-\frac1{n+1}\int_0^1\frac{1-t^{n+1}}{1-t}\,\mathrm{d}t
\text{(3e)}: \gamma=\lim\limits_{n\to\infty}H_n-\log(n)
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